Problem: Compute the value of $x$ such that

$\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\cdots\right)\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots\right)=1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\cdots$.
Explanation: The sum of an infinite geometric series with first term $a$ and common ratio $r$ is $\frac{a}{1-r}$.  Thus the sum of the first series is

$$\frac{1}{1-\frac{1}{2}}$$And the sum of the second series is

$$\frac{1}{1+\frac{1}{2}}$$Multiplying these, we get

$$\frac{1}{1-\left(\frac{1}{2}\right)^2}=\frac{1}{1-\frac{1}{4}}$$So $x=\boxed{4}$.